Divide the following complex numbers. $ \dfrac{-8-20i}{-2-5i}$
Solution: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-2+5i}$ $ \dfrac{-8-20i}{-2-5i} = \dfrac{-8-20i}{-2-5i} \cdot \dfrac{{-2+5i}}{{-2+5i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-8-20i) \cdot (-2+5i)} {(-2-5i) \cdot (-2+5i)} = \dfrac{(-8-20i) \cdot (-2+5i)} {(-2)^2 - (-5i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-8-20i) \cdot (-2+5i)} {(-2)^2 - (-5i)^2} = $ $ \dfrac{(-8-20i) \cdot (-2+5i)} {4 + 25} = $ $ \dfrac{(-8-20i) \cdot (-2+5i)} {29} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-8-20i}) \cdot ({-2+5i})} {29} = $ $ \dfrac{{-8} \cdot {(-2)} + {-20} \cdot {(-2) i} + {-8} \cdot {5 i} + {-20} \cdot {5 i^2}} {29} $ Evaluate each product of two numbers. $ \dfrac{16 + 40i - 40i - 100 i^2} {29} $ Finally, simplify the fraction. $ \dfrac{16 + 40i - 40i + 100} {29} = \dfrac{116 + 0i} {29} = 4 $